Centre of gravity of common plane laminas For which of the following planar shapes is the centre of gravity not at its simple geometric centre (i.e., not at a point of central symmetry)?

Introduction / Context:The centre of gravity (centroid) of symmetric shapes lies at their geometric centre. Shapes lacking central symmetry do not have a geometric centre coincident with the centroid. This concept is widely used in calculating bending stresses and deflections.

Given Data / Assumptions:

• Uniform thickness, homogeneous lamina.
• Plane shapes considered: circle, square, rectangle, equilateral triangle, right-angled triangle.

Concept / Approach:Shapes with full central symmetry (circle, square, rectangle) have the centroid at the intersection of their symmetry axes (geometric centre). An equilateral triangle has threefold symmetry; its centroid is at the intersection of medians, also the geometric centre of that symmetric triangle. A right-angled triangle lacks central symmetry; its centroid lies at the intersection of medians at distances b/3 and h/3 from the right-angled vertex, not at any “centre” of symmetry.

Step-by-Step Solution:

Verification / Alternative check:Compute centroid coordinates for a right-angled triangle with legs along axes: (x̄, ȳ) = (b/3, h/3) from the right-angle vertex. This is not the geometric centre of the bounding rectangle or of the triangle.

Why Other Options Are Wrong:

• Circle, square, rectangle: clear central symmetry → centroid at geometric centre.
• Equilateral triangle: high symmetry implies centroid is the intuitive centre point.

Common Pitfalls:Confusing “any triangle” with “equilateral triangle.” Only the equilateral has a central point that could be called a geometric centre coincident with the centroid.

Final Answer:Right-angled triangle